The hydrogen -like species `Li^(2+)` is in a spherically symmetric state `S_(1)` with one node. Upon absorbing light , the ion undergoes transition to a state `S_(2)`. The state `S_(2)` has one radial node and its energy is equal is to the ground state energy of the hydrogen atom.
The sate `S_(1)` is

Options

A. 1 s
B. 2 s
C. `2p`
D. 3 s

Explanation

To solve the problem, we need to determine the state \( S_1 \) of the hydrogen-like species \( \text{Li}^{2+} \) given the conditions in the question.

### Step-by-Step Solution:

1. **Understanding the Given Information**:
- The species is \( \text{Li}^{2+} \), which is a hydrogen-like ion.
- The state \( S_1 \) has one node and is described as being in a spherically symmetric state.

2. **Identifying the Spherical Symmetry**:
- A spherically symmetric state typically refers to an \( s \)-orbital, where the azimuthal quantum number \( l = 0 \).

3. **Understanding Nodes**:
- There are two types of nodes: radial nodes and angular nodes.
- The formula for the number of radial nodes is given by \( n - l - 1 \).
- The number of angular nodes is equal to \( l \).
- The total number of nodes is the sum of radial and angular nodes, which can be expressed as:
\[ \text{Total Nodes} = n - 1 \]

4. **Setting Up the Equation**:
- Given that \( S_1 \) has one node, we set up the equation:
\[ n - 1 = 1 \]
- Solving this gives:
\[ n = 2 \]

5. **Determining the Azimuthal Quantum Number**:
- Since \( S_1 \) is in a spherically symmetric state, we have:
\[ l = 0 \]
- This means that the orbital is an \( s \)-orbital.

6. **Identifying the State**:
- The principal quantum number \( n = 2 \) and the azimuthal quantum number \( l = 0 \) indicate that the state \( S_1 \) is:
\[ 2s \]

7. **Conclusion**:
- Therefore, the state \( S_1 \) is \( 2s \).

### Final Answer:
The state \( S_1 \) is \( 2s \).

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